The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $113$ years; the standard deviation is $14$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living longer than $99$ years.
$113$ $99$ $127$ $85$ $141$ $71$ $155$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $113$ years. We know the standard deviation is $14$ years, so one standard deviation below the mean is $99$ years and one standard deviation above the mean is $127$ years. Two standard deviations below the mean is $85$ years and two standard deviations above the mean is $141$ years. Three standard deviations below the mean is $71$ years and three standard deviations above the mean is $155$ years. We are interested in the probability of a turtle living longer than $99$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the turtles will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $99$ years and the other half $({16\%})$ will live longer than $127$ years. The probability of a particular turtle living longer than $99$ years is ${68\%} + {16\%}$, or $84\%$.